Question

altitude
Numericals :
Find the moment of force of 20 N about an a
of rotation at a distance of 0-5 m from the force
​

Answer 1

Given :

Moment of force of 20 N about rotation at a distance of 0.5 m from the force

To Find :

Moment of force

Solution :

Here moment of force is nothing but " Torque "

Torque :

It is defined as as product of force and perpendicular distance from point of rotation

Force , F = 20 N

Perpendicular distance , d = 0.5 m

$\sf \huge{\green{\tau=Fs}}\ \; \bigstar$

$:\implies \sf \tau =(20)(0.5)\\\\:\implies \sf \tau =(20)\left(\dfrac{5}{10}\right)\\\\:\implies \sf \tau =2(5)\\\\:\implies \sf \tau =10\ Nm$

Moment of force = 10 N-m

Answer 2

Answer:

✡ Given ✡

➡ Moment of force of 20 N about rotation at a distance of 0.5 m from the force.

✡ To Find ✡

➡ What is the moment of force.

✡ Formula Used ✡

⭐ $\tau$ = Force × Perpendicular distance ⭐

✡ Solution ✡

$\implies$ $\tau$ = 20 N × 0.5 m

$\implies$ $\tau$ = 20 N × $\dfrac{5}{10}$m

$\implies$ $\tau$ = 2(5) N.m

$\implies$ $\tau$ = 10 N.m

$\therefore$ Moment of force is 10 N.m

Explanation:

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